package main.leetcode.clockin.March;

/**
 * 999.车的可用捕获量
 *
 * <p>在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p”
 * 给出。大写字符表示白棋，小写字符表示黑棋。
 *
 * <p>车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。
 *
 * <p>返回车能够在一次移动中捕获到的卒的数量。
 *
 * <p>示例1：
 * 输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
 * 输出：3 解释： 在本例中，车能够捕获所有的卒。
 *
 * <p>示例2：
 * 输入：[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
 * 输出：0 解释： 象阻止了车捕获任何卒。
 *
 * <p>示例3：
 * 输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
 * 输出：3 解释： 车可以捕获位置 b5，d6 和 f5 的卒。
 *
 * <p>提示：board.length == board[i].length == 8 board[i][j] 可以是 'R'，'.'，'B' 或 'p'
 * 只有一个格子上存在 board[i][j] == 'R'
 *
 * <p>来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/available-captures-for-rook
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */
public class day26 {
    public static void main(String[] args) {
        System.out.println(
                new day26()
                        .numRookCaptures(
                                new char[][] {
                                    {'.', '.', '.', '.', '.', '.', '.', '.'},
                                    {'.', '.', '.', 'p', '.', '.', '.', '.'},
                                    {'.', '.', '.', 'p', '.', '.', '.', '.'},
                                    {'p', 'p', '.', 'R', '.', 'p', 'B', '.'},
                                    {'.', '.', '.', '.', '.', '.', '.', '.'},
                                    {'.', '.', '.', 'B', '.', '.', '.', '.'},
                                    {'.', '.', '.', 'p', '.', '.', '.', '.'},
                                    {'.', '.', '.', '.', '.', '.', '.', '.'}
                                }));
    }

    public int numRookCaptures(char[][] board) {
        for (int i = 0; i < 8; i++) {
            for (int j = 0; j < 8; j++) {
                if (board[i][j] == 'R') {
                    /* 模拟移动1 */
                    //                    return move(board, i, j, 1)
                    //                            + move(board, i, j, 2)
                    //                            + move(board, i, j, 3)
                    //                            + move(board, i, j, 4);

                    /* 模拟移动2 */
                    //                    return move(board, i, j);

                    /* 模拟移动3 */
                    return move(board, i, j, 1, 0)
                            + move(board, i, j, -1, 0)
                            + move(board, i, j, 0, -1)
                            + move(board, i, j, 0, 1);
                }
            }
        }
        return 0;
    }

    // 模拟移动3
    private int move(char[][] board, int x, int y, int dx, int dy) {
        while (x >= 0 && x < 8 && y >= 0 && y < 8 && board[x][y] != 'B') {
            if (board[x][y] == 'p') return 1;
            x += dx;
            y += dy;
        }
        return 0;
    }

    // 模拟移动2
    private int move(char[][] board, int x, int y) {
        int count = 0;
        int[] dx = {-1, 1, 0, 0};
        int[] dy = {0, 0, -1, 1};
        for (int i = 0; i < 4; i++) {
            for (int tx = x, ty = y;
                    tx >= 0 && tx < 8 && ty >= 0 && ty < 8;
                    tx += dx[i], ty += dy[i]) {
                if (board[tx][ty] == 'B') break;
                if (board[tx][ty] == 'p') {
                    ++count;
                    break;
                }
            }
        }
        return count;
    }

    // 模拟移动1
    private int move(char[][] board, int i, int j, int dir) {
        switch (dir) {
            case 1:
                while (--i > -1 && board[i][j] == '.') ;
                if (i == -1 || board[i][j] == 'B') return 0;
                else return 1;
            case 2:
                while (++i < 8 && board[i][j] == '.') ;
                if (i == 8 || board[i][j] == 'B') return 0;
                else return 1;
            case 3:
                while (--j > -1 && board[i][j] == '.') ;
                if (j == -1 || board[i][j] == 'B') return 0;
                else return 1;
            case 4:
                while (++j < 8 && board[i][j] == '.') ;
                if (j == 8 || board[i][j] == 'B') return 0;
                else return 1;
            default:
                return 0;
        }
    }
}
